Non-convex combination of vectors

Question

You are given a series of vectors $u_0,\ldots,u_k$ of non-negative entries, each of dimension $n$, for which the sum of the entries of each vector is $1$ (i.e., for all $i\in \{0,\ldots,k\}$, it holds that $\sum_j u_{i,j}=1$).

Is it always true that either there exists a vector $v\in R^{n}$, such that $\sum_j (u_{0,j}-u_{i,j})\cdot v_j > 0$ for all $i=1,\ldots,k$, or there exist $\alpha_1,\ldots,\alpha_k\geq 0$, that sum to $1$ (i.e., $\sum_i \alpha_i = 1$), such that $u_0 = \sum_{i=1}^k \alpha_i \cdot u_i$.

Answer 1

Yes, either point $u_0$ belongs to the convex hull $H$ of the points $u_1,\dots,u_k$ (in which case $\alpha_i$ exist), or there exists a hyperplane separating $u_0$ and $H$, when we can take $v$ as a normal vector of this hyperplane "pointing" towards $u_0$.